In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed

Question

mean of 55.2 inches, and standard deviation of 2.5 inches.

A) What is the probability that a randomly chosen child has a height of less than 59.55 inches?

B) What is the probability that a randomly chosen child has a height of more than 52.7 inches?

Answer 1

Hi Bailey,

This is our old friendly formula z=(x-mu)/sigma

a) x=value you were given

Mu=mean

Sigma=standard deviation

Z=z-score

For our problem:

x=59.55

mu=55.2

sigma=2.5

Thus:

z=(59.55-55.2)/2.5

z=1.74

From z-table:

P(Z<1.74)=P(X<59.55)=0.9591

b) New value for x, rest stay the same.

x=52.7

mu=55.2

sigma=2.5

z=z-score

z=(52.7-55.2)/2.5

z= -1.00

From z-table:

P(Z<-1.00)=P(X<52.7)=0.1587

But we want greater than 52.7, so:

P(Z> -1.00)=P(X>52.7)= 1-P(X<52.7)=

1-0.1587=

P=0.8413

I would suggest looking over the worked solutions other tutors and I have posted here, maybe do something fun, then come back and see if you can work the problems on your own with our solutions as an answer key. Good luck.